Light of wavelength 300 nm in medium A enters into medium B through a plane surface. If the frequency of light is 5×10^14 Hz and the ratio of speed in medium A to that is medium B is 4/5, the absolute refractive index of medium B is,

Correct answer: (A) 1.6. Absolute refractive index of medium A, μ_A=c/(ν_Aλ)=(3×10^8)/(5×10^14×300×10^(-9))=2 Refractive index is inversely proportional to the velocity of light. ⇒μ_Bμ_A=v_Av_B=4/5 ⇒μ_B=μ_A4/5=8/5 ⇒μ_B=1.6

Q145 / 160 · AP EAMCET 2021 (19 Aug Shift 2) · Physics

Light of wavelength $300 nm$ in medium $A$ enters into medium $B$ through a plane surface. If the frequency of light is $5 \times 10^{14} Hz$ and the ratio of speed in medium $A$ to that is medium $B$ is $\frac{4}{5}$ , the absolute refractive index of medium $B$ is,

Correct answer
A
$1.6$
B
$1.5$
C
$1.3$
D
$1.45$

Solution

Absolute refractive index of medium A, $μ_{A} = \frac{c}{ν_{A} λ} = \frac{3 \times 10^{8}}{5 \times 10^{14} \times 300 \times 10^{- 9}} = 2$

Refractive index is inversely proportional to the velocity of light.

$\Rightarrow \frac{μ_{B}}{μ_{A}} = \frac{v_{A}}{v_{B}} = \frac{4}{5}$

$\Rightarrow μ_{B} = μ_{A} \frac{4}{5} = \frac{8}{5}$

$\Rightarrow μ_{B} = 1.6$

Hence, the correct answer is option: 'A'.

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Light of wavelength 300 nm in medium A enters into medium B through a plane surface. If the frequency of light is 5×1014 Hz and the ratio of speed in medium A to that is medium B is 45, the absolute refractive index of medium B is,

Solution

Relevant Questions

Light of wavelength $300 nm$ in medium $A$ enters into medium $B$ through a plane surface. If the frequency of light is $5 \times 10^{14} Hz$ and the ratio of speed in medium $A$ to that is medium $B$ is $\frac{4}{5}$ , the absolute refractive index of medium $B$ is,